Why are aftershocks dangerous to buildings after a large earthquake

An electron that has moved from the outer energy levels towards the inner energy levels of an atom is said to be in a ground sta

IRISSAK [1]

I think it’s true....

7 0

2 years ago

Which statement is NOT true?

Nuetrik [128]

Answer:

light doesn't need a medium through which to travel because the speed of light is experimentally constant

4 0

2 years ago

PLEASE HELP ASAP

Dmitrij [34]

The answer should be B

3 0

3 years ago

Read 2 more answers

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

3 years ago

malcom is doing a wheely on his motorcycle and moving at 20 m/s with a momentum of 6000 kg m/s when he lays her down. The mass o

rodikova [14]

Answer:

110 kg

Explanation:

The momentum of the Malcom+motorcycle system is given by:

$p=(m+M)v$

where

(m+M) is the total mass of the system, with

m = mass of Malcom

M = 190 kg (mass of the motorcycle)

v = 20 m/s velocity

Since we know the momentum:

p = 6000 kg m/s

We can re-arrange the equation to find the mass of Malcom:

$m=\frac{p}{v}-M=\frac{6000 kg m/s}{20 m/s}-190 kg=110 kg$

3 0

3 years ago